Unformatted text preview: MA294: A Sample Correction Homework # 2
Acmae El Yacoubi September 5, 2007
ex. 2.2.2 Given the relation (2.2.3, p. 63), the matrix T is
T = cos(60 )  sin(60 ) sin(60 ) cos(60 ) =
1 2 3 2  1 2 3 2 ex. 2.2.10 The unit vector on the line L is
u= 1 4 5 3 For any x R2 where x = [x1, x2], we have:
projL (x) = (u.x)u = ( 4 x1 + 3 x2 ) 5 5 = = 4/5 3/5 0.64x1 + 0.48x2 0.48x1 + 0.36x2 0.64 0.48 x . 1 x2 0.48 0.36 This being true for all x, we nally obtain the matrix of the projection onto the line L
A= 0.64 0.48 0.48 0.36 ex. 2.2.16 Let T be the reection about a line L in R2 where L is shown below 1 L a. b. Reect the red vector about the line L. By denition, the matrix of T is obtained by mapping the basis vectors, namely e1 and e2. That is, [T (e1) T (e2)]. Given the denition of the reection (def. 2.2.2, p 60), we have:
T (x) = 2projL (x)  x = 2(u.x)u  x where u is the unit vector on L. For our case
u= cos() sin() 1 Thus
T (e1 ) = 2(u.e1 )u  e1 = = cos(2) sin(2) 2 cos2 ()  1 2 cos() sin() and
T (e2 ) = 2(u.e2 )u  e2 = = sin(2)  cos(2) 2 cos() sin() 2 sin2 ()  1 Finally, the matrix T is
T () = cos(2) sin(2) sin(2)  cos(2) ex. 2.2.22 Let e1 , e2 , e3 be the unit vectors on the xaxis, yaxis and z axis, respectively. The counterclockwise rotation about the yaxis, noted T , acts on the unit vectors as follows:
T (e1 ) = cos()e1  sin()e3 T (e2 ) = e2 T (e3 ) = sin()e1 + cos()e3 cos() 0 sin() 1 0 T = 0  sin() 0 cos() Thus its matrix is ex. 2.2.42 T being the projection onto the line L. for any vector x we have
T (x) = projL (x) = (u.x)u where u is the unit vector on L. Thus, since the vector u is on the line L, it is invarient by the projection T , i.e T (u) = u. We can show this result by applying T twice onto x as follows
T (T (x)) = T ((u.x)u) = (u.x)u = T (x) where we used the linearity of T . This being true for all x R2 , we nally have
T2 = T Note: We say that T is idempotent.
ex. 2.3.14 The square matrix A is dened by 2 1 0 0 5 3 0 0 0 0 1 2 0 0 2 5 0 0 0 1 To check if it is invertible, we write the augmented matrix [AI4 ] and we compute its rref as follows
2 1 0 0 5 3 0 0 0 0 1 2 0 0 2 5     1 0 0 0 0 1 0 0 0 1 5 0 0  2 0 1 0 0  0 2 0 0 1 2  0 0 0 2 5  1 0 0 1 0 0 0  0 1 0 0  0 0 0 0 1 0  1 0 2 1 0 0 0 1  0 0 1 0 1 2 0 0
1 2 1 5 0 0 2 0 1 0 0 0 0 1 2 0 0 0 1  1 0 2  1 2  0 0  0 0 3 5 0 0 1 2 0 0 0 0 5 2 0 0 2 1 0 1 0 0 0 0 1 0 2 [I4 A1 ] Thus, A is invertible and its inverse is
A1 3 5 0 0 1 2 0 0 = 0 0 5 2 0 0 2 1 Note: For your information, A is a diagonal blockmatrix. In this case, the inverse can easily be found by computing the respective inverse of the 2 2 submatrices, A1 and A2 :
A1 = A1 O2 O2 A2 1 2 2 5 0 0 0 0 where
A1 = 2 5 1 3 A2 = O2 = ex. 2.3.34 Let A be the matrix dened by a 0 0 A = 0 b 0 0 0 c a. A is invertible i det(A) = 0. Since A is diagonal, it is easy to compute its determinant: it's the product of its diagonal entries. Thus
det(A) = 0 abc = 0 (a = 0 and b = 0 and c = 0) If det(A) = 0, then its inverse is 1 A1 = 0 0 a 0 0
1 b 0 0
1 c b. The above mentioned rule can be applied to diagonal matrices of any size n. That is, a diagonal matrix A = [aij ] of order n is invertible if and only if
n aii = 0
i=1 ex. 2.3.40 Let A = [C1 , C2 , ..., Cn ] be a matrix of order n such that (i, j) , i = j : Ci = Cj i.e. A has two columns that are equal. Assume that T is the linear transformation associated with A and that A is represented w.r.t a basis e = {e1 , e2 , ..., en }. We know that each column Ck of A is the image a basis vector ek (see Fact 2.1.2, p 48) in the domain of T . Thus
Ci = Cj T (ei ) = T (ej ) T (ei  ej ) = 0 u = 0, T (u) = 0 (u = ei  ej ) T (thus A) is non invertible Q.E.D ex. 2.3.42 Permutation matrices A permutation matrix has a '1' exactly once in each row and column. By swapping rows, we end up having the identity matrix with leading 1's in each row. Thus, a permutation matrix is invertible. 3 Let be a permutation of {1, 2, ..., n} and P the associated permutation matrix dened by 1 1 P = [i(j) ]. Its inverse , P = P , is also a permutation matrix. 1 ( ) You can easily check that by computing the ith rowj th column of P P .
ex. 2.3.52 Let b R4 such that Ax = b is inconsistent. The matrix representation of this linear system is: 0 0 0 1 1 2 3 4 2 4 6 8     b1 b2 b3 b4  b4  b1  b2  2b1  b3  3b1 We compute the rref of [Ab ]:
1 0 0 0 4 1 2 3 8 2 4 6     b4 1 4 0 1 b1  0 0 b2 0 0 b3 8 2 0 0 The third and last row implies that (b2  2b1 = 0 and b3  3b1 = 0). Thus, to have an inconsistent system, we only need to make sure that either one of these two conditions is not met. For example, we can choose: b1 = 1, b2 = 0 and set anything for b3 and b4 . Finally, an example of b could be 1 0 b= 0 1 ex. 2.4 16  26 16 True. 17 False. (A + B)2 = A2 + AB + BA + B 2 . The matrix product is not commutative. 18 True. 19 False. The inverse is not distributive. 20 False. (A  B)(A + B) = A2 + AB  BA + B 2 . The matrix product is not commutative. 21 True. 22 False. The matrix product is not commutative. 23 True. (ABA1 )3 = ABA1 ABA1 ABA1 = AB 3 A1 . 24 True. 25 True. ex. 2.4.26 Block matrices Let us write 1 0 A= 0 0 1 3 B= 0 0 0 1 0 0 2 4 0 0 1 0 1 0 2 4 1 3 0 1 = A1 0 A3 1 3 5 = B1 2 B3 4 A2 A4 B2 B4 where A1 = A2 = A4 = I2 , A3 = B3 = O2 . To compute the product AB , we multiply the corresponding Ai and Bj 2 2 matrices as follows: A1 B1 + A2 B3 A1 B2 + A2 B4 AB = A3 B1 + A4 B3 A3 B2 + A4 B4 4 Plugging the expression of the products of the submatrices Ai Bj , we obtain
1 3 AB = 0 0 2 4 0 0 2 4 1 3 3 5 2 4 ex. 2.4.34 We have
AB(AB)1 = In = A.(B(AB)1 ) = AC where C = B(AB)1 A Thus, according to (fact 2.4.9, p 85): Likewise, we have and C are both invertible and A1 = C = B(AB)1 (AB)1 AB = In = ((AB)1 A).B = DB where D = (AB)1 A Thus, according to (fact 2.4.9, p 85): B and D are both invertible and B 1 = D = (AB)1 A ex. 2.4.40 We have
B 1 = 1 2 3 5 and (AB)1 = 1 3 2 5 and (see ex. 2.4.34) Thus, if we write (AB)1 AB = In A= = a b c d B 1 = (AB)1 A and plug it into the previous equation, we obtain
B 1 = 1 2 3 5 = 1 3 a b . 2 5 c d = a + 3c b + 3d 2a + 5c 2b + 5d We have 2 2 linear systems to solve. We nally obtain
a = 4, b = 5, c = 1. d = 1 A= 4 5 1 1 ex. 2.4.42 a. The angle between the lines P and Q is = 30 . Let uP and uQ Q, respectively. For a given vector x, we use the following notations
(x, uP ) = and (ref P (x), T (x)) = be the unit vectors on P and Note that + = = 30 . Thus, the angle between x and T (x) is
(x, T (x)) = = = i.e. (x, T (x)) = (x, ref P (x)) + (ref P (x), T (x)) 2 (x, uP ) + 2 (refP (x), uQ ) 2+2 60 Since the reection does not change the length, x and T (x) have the same length. b. From (a), we deduce that: 5 x and T (x) have the same length, and the angle between these two vectors is = 2
Thus, T is the rotation of angle = 2 (counterclockwise). c. From (b) and (fact 2.2.3, p 63), we conclude that T = cos()  sin() sin() cos() = 1 2 3 2  1 2 3 2 ex. 2.4.50 a. Let A and E be two matrices dened by a b c R1 d e f = R2 E = A= R3 g h k
The product EA is 1 0 0 3 1 0 0 0 1 a b c R1 EA = d  3a e  3b f  3c = R2 3R1 R3 g h k 0 0 0 = 1 0 0 0 0 0 Note that E = I3  3.E21 where E21 E is an elementary matrix which adds (3)times the 1st row of A to its second row. i.e. R2  R2 + (3)R1 b. Let A be an arbitrary matrix dened by its rows and E be the matrix dened as follows 1 0 0 R1 0 1 0 R 2 E= A= 4 R3 0 0 1
The product EA is R1 EA = 1 R2 4 R3 Hence, EA is obtained from A by multiplying this latter's 2nd row, R2 , by a scalar (1/4). E is an elementary matrix. c. We want to swap the two last rows of a 3 3matrix A. Consider the elementary matrix E dened by 1 0 0 E = 0 0 1 0 1 0 E is obtained from the identity matrix I3 check that 1 EA = 0 0 by swapping this latter's last two rows. We can easily 0 0 R1 R1 0 1 . R2 = R3 R3 R2 1 0 d. In (a), E is obtained from the identity matrix I3 by adding (3) times this latter's rst row to its second row. In (b), E is obtained from the identity matrix I3 by multiplying this latter's second row by a scalar. In (c), E is obtained from the identity matrix I3 by swapping this latter's last two rows.
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 Fall '05
 HUI
 Linear Algebra, Algebra, Matrices, Diagonal matrix, Det

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